REVIEW OF CHAPTERS 8-9
1.
b.
3.
c.
5.
d.
7.
a.
9.
a, b, c.
11.
Since the population distribution is approximately normal and the sample size (n = 35) is
reasonably large, the distribution of sample mean, X , is approximately normal with mean
 X    8.6 , and standard deviation,  X 

n

2.0
 0.388 .
35
That is, X ~ Normal (8.6, 0.338) .
We want probability that value of X will be at least 8.8.
z-value corresponding to x = 8.8 is z =
8.8  8.6
 0.592 .
0.338
Using the Z-table, we find that the area under the Z-curve to the right of 0.592 is
approximately (0.5 – 0.223) = 0.277.
Thus, the likelihood that the mean of the sample is at least 8.8 weeks is approximately
0.277.
13.
Since the population distribution is approximately normal and the sample size (n = 64) is
large enough, the distribution of
X 
is approximately student’s t-distribution with df =
S n
n – 1 = 63.
(1 - ) = 0.98. Hence,  = (1 – 0.98) = 0.02. From the t-table, we find that for df = 63, t/2
= t0.01 = 2.381.
A 98 percent confidence interval estimate for mean number of customers per day

is  x  t0.01

s
s  
20
20 
, x  t0.01
,160  2.381
 = 160  2.381

n
n 
64
64 
= (154.0475, 165.9525).
15.
Since the population distribution is approximately normal, the distribution of
approximately student’s t-distribution with df = n – 1.
For the given sample data,
x
s
826  931 
6
 1101
= 985.5
(829  985.5) 2  (931  985.5) 2 
5
 (1101  985.5) 2
= 115.4968.
X 
is
S n
(1 - ) = 0.95. Hence,  = (1 – 0.95) = 0.05. Using the t-table, we find that for df = n – 1 =
5, t/2 = t0.025 = 2.015.
Hence, a 95 percent confidence interval estimate for the population mean, , is
s 

 x  t0.025
=
n

17.
115.4968 

 985.5  2.015
 = (890.49, 1080.51).
6 

Since the population distribution is approximately normal, the distribution of
X 
is
S n
approximately student’s t-distribution with degree of freedom df = n – 1 = 15.
(1 - ) = 0.95. Hence,  = (1 – 0.95) = 0.05. Using the t-table, we find that for df = n – 1 =
15, t/2 = t0.025 = 2.131.
Hence, a 95 percent confidence interval estimate for the population mean, , is
s 

 x  t0.025
=
n

35 

 240  2.131
 = (221.3538, 258.6463).
16 

The value 250 lies inside the interval estimate obtained. The analysis therefore does not
provide significant evidence to conclude that the mean daily production of widgets has
increased.
Since the population distribution is approximately normal, the distribution of X is
approximately normal.
19.
The width  E of 95 percent confidence interval is  z0.05

n
 1.96
25
. We want the
n
width should be no more than 4.0.
 z    1.96  25 
Hence, the sample size should be at least  0.05  = 
 = 150.06 or 151.
4
 E  

2
21.
2
Let us hope that the required sample size will be large enough to use normal
approximation.
Thus, using normal approximation and the estimate p = 0.08 of p, we get the following
approximate lower bound on the sample size n.
An approximate lower bound for the sample size is
2
2
 z0.05 p 1  p  
  2.326   0.08 0.92  

 = 
  995.5 or 996.


E
0.02






996 (0.08) > 5 and 996 (0.92) > 5.
Hence, this sample size is large enough for normal approximation as hoped. The bound is
then valid.
We should choose a sample of size at least 996.
23.
Let p be the fraction of Canadian adults who believe that government of Canada should
give a high priority to the issue of Canadian unity.
The point estimate of p, obtained from the sample is p̂ = 0.62.
(a)
n p̂ = (4704) (0.62) > 5 and n (1 - p̂ ) = (4704) (0.38) > 5.
Thus, the sample size n is large enough for use of normal approximation.
(1 - ) = 0.95. So,  = 0.05. z/2 = z0.025 = 1.96.
An approximately 95 percent confidence interval estimate for p is
pˆ  z0.025
(b)
  0.62  0.38 
pˆ (1  pˆ )

= 0.62  1.96 


4704
n


=  0.62  0.014 = (0.606, 0.634).
The entire confidence interval estimate obtained in part (a) lies above 0.6. Hence, it will be
reasonable to conclude that the statement that “p is less than 0.6” is incorrect.